• Radius (r): Constant distance from the center of the circle to the object.
• Velocity (v): Always tangent to the circle, constant in magnitude.
• Centripetal force (F_c): Directed toward the center, keeps the object moving in a circle.

$F_c = \frac{mv^2}{r}$

• Angular velocity (ω): Rate of change of angular displacement.
• Relationship between linear and angular velocity:

$v = r\omega$

• Period (T): Time for one full revolution.
• Frequency (f): Number of revolutions per second.

$T = \frac{2\pi}{\omega}, \quad f = \frac{1}{T}$

Uniform circular motion refers to motion of a body along a circular path with constant speed. While the speed remains constant, the direction of the velocity vector changes continuously. This requires a net force directed toward the center of the circle, known as centripetal force.

a. Angular displacement

It is the angle turned by a body moving along a circular path: $\theta = \frac{l}{r}$ where:

• $\theta$ is angular displacement in radians
• $l$ is arc length
• $r$ is radius

b. Angular velocity

Angular velocity is the rate of change of angular displacement: $\omega = \frac{d\theta}{dt}$ If the object completes a full revolution in time $T$: $\omega = \frac{2\pi}{T}$

Linear velocity is related to angular velocity by: $v = r\omega$

c. Centripetal acceleration

This is the acceleration that keeps an object moving in a circle: $a_c = \frac{v^2}{r}$ Alternatively: $a_c = r\omega^2$

This is the net force causing centripetal force: $F_c = \frac{mv^2}{r} = mr\omega^2$

When the speed of an object in circular motion changes, it has a tangential acceleration: $a_t = \frac{dv}{dt} = r\alpha$ where $\alpha$ is angular acceleration.

The net acceleration is the vector sum of centripetal and tangential acceleration: $a = \sqrt{a_c^2 + a_t^2}$

Consider an object of mass $m$ tied to a string and rotated in a circular path on a horizontal plane. When moving at constant speed, the string maintains an angle $\theta$ with the vertical. This forms a conical pendulum.

Force analysis

There are two components of tension $T$ in the string:

• Vertical component: $T\cos\theta = mg$ (Equation 3.11)
• Horizontal component: $T\sin\theta = \frac{mv^2}{r}$ (Equation 3.10)

Dividing the two equations:

$\frac{T\sin\theta}{T\cos\theta} = \frac{mv^2 / r}{mg} \Rightarrow \tan\theta = \frac{v^2}{rg}$

Also, since $v^2 = \omega^2 r^2$, we have:

$\tan\theta = \frac{\omega^2 r}{g}$

Conical pendulum period

Let $l$ be the length of the string. Then:

• Radius of circular motion: $r = l\sin\theta$
• Vertical component: $l\cos\theta = \frac{gT^2}{4\pi^2} \Rightarrow T = 2\pi \sqrt{\frac{l\cos\theta}{g}}$

Example calculation

Given:

• Mass $m = 0.5\,\text{kg}$
• Radius $r = 1\,\text{m}$
• Speed $v = 10\,\text{ms}^{-1}$

a. Centripetal acceleration:

$a_c = \frac{v^2}{r} = \frac{(10)^2}{1} = 100\,\text{ms}^{-2}$

b. Centripetal force:

$F_c = \frac{mv^2}{r} = \frac{0.5 \times 100}{1} = 50\,\text{N}$

Therefore, the centripetal acceleration is $100\,\text{ms}^{-2}$ and the force acting on the stone is $50\,\text{N}$.

Consider an object of mass $m$ tied to a string and whirled in a circular path on a horizontal plane, forming a conical pendulum.

Forces in the conical pendulum

• Vertical force balance: $T\cos\theta = mg$
• Horizontal component (centripetal force): $T\sin\theta = \frac{mv^2}{r}$
• Dividing the equations: $\tan\theta = \frac{v^2}{rg}$
• Since $v = \omega r$, then: $\tan\theta = \frac{\omega^2 r}{g}$

Example 1

Given: $m = 0.5\,\text{kg}, v = 10\,\text{ms}^{-1}, r = 1\,\text{m}$

1. Centripetal acceleration: $a_c = \frac{v^2}{r} = \frac{(10)^2}{1} = 100\,\text{ms}^{-2}$
2. Centripetal force: $F_c = \frac{mv^2}{r} = 0.5 \times 100 = 50\,\text{N}$

Example 2

Using $v = \omega r = 2\pi f r$, then: Tension: $T = 4\pi^2 f^2 m r$

Example 3 - Conical pendulum

Given: $l = 1\,\text{m}, r = 0.6\,\text{m}, m = 1\,\text{kg}$

1. Find $\cos\theta = \frac{\sqrt{l^2 - r^2}}{l} = \frac{\sqrt{1^2 - 0.6^2}}{1} = 0.8$
2. Tension: $F = \frac{mg}{\cos\theta} = \frac{1 \times 9.8}{0.8} = 12.25\,\text{N}$
3. Period: $T = 2\pi \sqrt{\frac{l \cos\theta}{g}} = 2\pi \sqrt{\frac{1 \times 0.8}{9.8}} \approx 1.79\,\text{s}$

Summary of key equations

• Centripetal acceleration: $a_c = \frac{v^2}{r} = \omega^2 r$
• Centripetal force: $F_c = \frac{mv^2}{r} = m\omega^2 r$
• Vertical balance: $T\cos\theta = mg$
• Horizontal force: $T\sin\theta = \frac{mv^2}{r}$
• Period of conical pendulum: $T = 2\pi \sqrt{\frac{l\cos\theta}{g}}$

When a vehicle turns in a circular path, it behaves differently when on a flat and a banked road. The centripetal and frictional forces acting on the vehicle determine the maximum safety velocity the vehicle can travel. On a flat surface, cars must rely only on friction to prevent skidding. During the rainy season, friction is reduced, hence the turning force becomes smaller. Therefore, banked curves are introduced to prevent skidding. With banked curves, the normal force provides a component of force directing a vehicle towards the centre of the curve, reducing the vehicle's dependency on friction alone.

a. A car on a level rough curved road

The necessary centripetal force is provided by the frictional force between the tyres of the car and the road. Let $R_1$ and $R_2$ be normal reactions of the inner and outer pairs of wheels respectively, and let $f_1$ and $f_2$ be the corresponding frictional forces. Let $h$ be the height of the centre of gravity above the ground, $a$ the distance between the car wheels, and $r$ the radius of the circular path.

Vertical forces: $R_1 + R_2 = mg$

Horizontal forces: $f_1 + f_2 = \frac{mv^2}{r}$

Taking moments about G:

$(f_1 + f_2)h = \frac{a}{2}(R_2 - R_1)$

Substituting equations into one another gives:

$R_1 = \frac{mg}{2} - \frac{hv^2}{ar}$

$R_2 = \frac{mg}{2} + \frac{hv^2}{ar}$

Maximum speed without toppling:

$v_c = \sqrt{\frac{arg}{2h}}$

If friction coefficient $\mu$:

$v_c = \sqrt{\mu rg}$

b. A car on a banked rough curved road

Let the banking angle be $\theta$. Let $R$ be the normal reaction, and $f$ the frictional force. Then:

Horizontal forces:

$R\sin\theta + f\cos\theta = \frac{mv_{\text{max}}^2}{r}$

Vertical forces:

$R\cos\theta = f\sin\theta + mg$

Solving the system yields:

$v_{\text{max}} = \sqrt{ \frac{rg(\tan\theta + \mu)}{1 - \mu\tan\theta} }$

$v_{\text{min}} = \sqrt{ \frac{rg(\tan\theta - \mu)}{1 + \mu\tan\theta} }$

c. A car on a banked smooth curved road

Here, no friction is involved. The normal reaction provides all the centripetal force.

Vertical forces:

$R\cos\theta = mg$

Horizontal forces:

$R\sin\theta = \frac{mv^2}{r}$

Dividing gives:

$v = \sqrt{rg\tan\theta}$

d. Cyclist on a curved rough level road

A cyclist must lean inward to create enough frictional force for centripetal acceleration.

Using the principle of moments:

$f \cdot h = R \cdot a$

Also, $\mu = \tan\theta$ and $v_c = \sqrt{rg\tan\theta}$

Example: Banking angle for safety

Given:

• Speed $v = 72\text{ km/h} = 20\text{ m/s}$
• Radius $r = 50\text{ m}$
• $g = 9.8\text{ m/s}^2$

$\tan\theta = \frac{v^2}{rg} = \frac{400}{490} = 0.8163$

$\theta = \tan^{-1}(0.8163) \approx 39^\circ$

The road should be banked at an angle of $39^\circ$.

There are several applications of circular motion in daily life; these include the following:

a. Rotating fluids

When a liquid in a container is stirred, the centre of the liquid surface forms a hollow. The surface of the liquid will be defined by how the centripetal acceleration changes with radius. A parabolic surface of the liquid may be used in liquid mirror telescopes. The most common liquid used is mercury (or low melting alloys of gallium). In these telescopes, the liquid and its container are rotated at a constant speed around a vertical axis, which causes the surface of the liquid to assume a "paraboloidal" shape regardless of the container's shape.

b. Centrifugal pump

The main part of a centrifugal pump is the impeller which has a series of curved vanes fitted inside the shroud plates. When a fluid (e.g. water) enters the pump along or near the rotating axis, it is accelerated by the pump impeller. The fluid particles then accelerate radially outward into a volute chamber (casting) from where it exits.

c. Centrifuge

A centrifuge is a device that is used for separating mixtures. It can be used to separate sugar crystals from molasses, cream from milk, bee wax from honey, and constituents of blood and urine samples.

The centrifuge works using the sedimentation principle. The sample of liquid mixture is spun at relatively high speed, creating a strong centripetal force on the liquid and its contents. This force will make denser particles accelerate outward in the radial direction. When the centrifuge is settled, the heavier (denser) particles settle to the bottom while lighter (less dense) particles rise to the top.